Which of the following statements is false?

Options

A. Permanent magnetic moment of `[Cu(NH_(3))_(4)]^(2+)` is 1.732 B.M.
B. Equilibrium constant is the ratio of rate constants of forward and backward reactions
C. `[Ni(CO_(4)]` is tetrahedral
D. For forming `NCl_(5)'N'` adopts `sp^(3)d` hybrid state

Answer

To determine which of the statements is false, we will analyze each statement one by one.

### Step 1: Analyze the first statement
**Statement:** The paramagnetic permanent magnetic moment of Cu(NH3)4^2+ is 1.732 Bm.

- **Analysis:**
- Copper (Cu) in the +2 oxidation state has an electronic configuration of 3d^9.
- In the presence of NH3, which is a strong field ligand, the electron configuration can lead to pairing of electrons.
- The number of unpaired electrons (N) in Cu(NH3)4^2+ is 1.
- The magnetic moment (μ) can be calculated using the formula:
\[ \mu = \sqrt{N(N + 2)} \]
\[ \mu = \sqrt{1(1 + 2)} = \sqrt{3} \]
- This gives us a magnetic moment of approximately 1.732 Bm.
- **Conclusion:** This statement is true.

### Step 2: Analyze the second statement
**Statement:** The equilibrium constant is the ratio of rate constants of forward and backward reactions.

- **Analysis:**
- For a reaction at equilibrium, the equilibrium constant (K_eq) is defined as the ratio of the concentration of products to reactants.
- The relationship between the equilibrium constant and the rate constants of the forward (K_f) and backward (K_b) reactions is given by:
\[ K_{eq} = \frac{K_f}{K_b} \]
- **Conclusion:** This statement is true.

### Step 3: Analyze the third statement
**Statement:** Ni(CO)4 is tetrahedral.

- **Analysis:**
- Nickel (Ni) has an atomic number of 28, with an electronic configuration of 3d^8 4s^2.
- In Ni(CO)4, CO is a strong field ligand and causes pairing of the d-electrons.
- The hybridization in this case is sp^3, which results in a tetrahedral geometry.
- **Conclusion:** This statement is true.

### Step 4: Analyze the fourth statement
**Statement:** For forming NCl5, N adopts an sp^3d hybrid state.

- **Analysis:**
- Nitrogen (N) has an atomic number of 7, with an electronic configuration of 2s^2 2p^3.
- Nitrogen does not have d-orbitals available for hybridization because the d-orbitals start from the third period.
- Therefore, nitrogen cannot form sp^3d hybridization.
- **Conclusion:** This statement is false.

### Final Conclusion:
The false statement among the options provided is the last one: "For forming NCl5, N adopts an sp^3d hybrid state."

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